Is a cap an Alexandrov space?

Question

Let us call a cap the intersection of the boundary of 3-dimensional convex compact set $K$ in $\mathbb{R}^3$ with a half-space bounded by a plane $H$ such that the orthogonal projection to $H$ of this intersection is contained in $K\cap H$.

Question. Is it true that a cap equipped with the intrinsic metric is an Alexandrov space (with boundary) with non-negative curvature.

Remark. In the case when $K$ has smooth boundary and $H$ intersects $K$ transversally, the question is equivalent to asking whether the second fundamental form of the boundary of the cap is non-negative.

Answer 1

Note that the union of $K\cap H$ with its reflection is a convex set. Therefore, its surface $\Sigma$ is an Alexandrov space. Your space is a quotient $\Sigma/\mathbb{Z}_2$ by isometric involution. Therefore, it is an Alexandrov space as well.

(There is a closely related problem Convex hat, page 21 in my PIGTIKAL.)